# Are Brown representable functors determined by restriction to finite complexes?

Assume two $CW$ complexes $X,Y$ give two functors $h_X=[-,X], h_Y=[-,Y]$ on the homotopy category of $CW$ complexes whose restrictions to the full subcategory of finite $CW$ complexes are naturally equivalent. Does this imply that the two functors are naturally equivalent (i.e. that X, Y are homotopy equivalent)?

• The main question is if natural equivalence of functors is represented by a map $X\to Y$. If so, then the spaces are weakly equivalent because this is tested on spheres, which are finite complexes. By Whitehead's theorem, they are also homotopy equivalent. Maybe the representability of the natural equivalence is discussed in Brown's paper? – Matthias Wendt Oct 3 '14 at 6:54

In particular, finite CW-complexes do not suffice since a map inducing bijections on homotopy classes of maps out of CW-complexes might not be a $\pi_1$-isomorphism. In general, it will only induce a bijection on conjugacy classes of $\pi_1$s. On the other hand, this answer shows that this works for maps between CW-complexes with finitely generated fundamental groups.
• Thanks for an interesting answer. Could you please comment on why this is true in the category of based connected CW complexes? Note that I don't assume the existence of a map $X \to Y$ inducing the equivalence of the restriction to finite complexes. – user58951 Oct 3 '14 at 14:38